A precipitate forms from a double displacement reaction or metathesis.
Explanation:
If given two clear solutions and upon reacting a cloudy/insoluble substance forms with the rest of the liquid being clear, a double displacement reaction has been carried out. The insoluble cloudy substance is called precipitate.
- the driving force for the bulk of double displacement reactions is the formation of precipitates.
- from careful observations, a solubility chart has been developed. If the compounds reacting are known, using the chart, a chemist can predict whether a precipitate will form or not.
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Im not sure but i think its hbr. i hope im right and i hope this helped,
Answer:
5.67 g OF WATER WILL BE FORMED WHEN 13.7 g OF MnO2 REACTS WITH HCl GAS.
Explanation:
EQUATION FOR THE REACTION
Mn02 + 4HCl --------> MnCl2 + Cl2 + 2H2O
From the balanced reaction between manganese oxide and hydrogen chloride gas;
1 mole of MnO2 reacts to form 2 mole of water
At STP, the molecular mass of the sample is equal to the mole of the substance. So therefore:
(55 + 16 * 2) g of MnO2 reacts to form 2 * ( 1 *2 + 16) g of water
(55 + 32) g of MnO2 reacts to form 2 * 18 g of water
87 g of MnO2 reacts to form 36 g of water
If 13.7 g of MnO2 were to be used?
87 g of MnO2 = 36 g of H2O
13.7 g of MnO2 = ( 13.7 * 36 / 87) g of water
= 493.2 / 87 g of water
Mass of water = 5.669 g of water
Approximately 5.67 g of water will be formed when 13.7 g of manganese oxide reacts with excess hydrogen chloride gas.
<span>The electronic configuration of cobalt is :1s2 2s2 2p6 3s2 3p6 3d7 4s2 </span>
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
